main() { int a; scanf(%d,&a); if(a%3==0&&a%5==0&&a%7==0) { printf("000"); } if(a%3==0&&a%5==0&&a%7!

Question

欢迎各位前来检查错误，谢谢。
程序设计 输入一个整数，判断它能否被 3,5,7整除，并输出以下信息之一：
a。能同时被3,5,7整除
b。能被其中两数（要指出那两个）整除：
c.能被其中一个数（要指出）整除：
d.不能被3,5,7中任一个整除。
针对我这个题目，还有更简单的编译方法吗？谢谢你们

Answer 1

#include <stdio.h>

int main()
{ int a;
scanf("%d",&a);
if(a%3==0&&a%5==0&&a%7==0)
{
printf("000");
}
if((a%3==0&&a%5==0&&a%7!=0)||(a%3==0&&a%7==0&&a%5!=0)||(a%5==0&&a%7==0&&a%3!=0))
{
printf("1111");
}
if((a%3==0&&a%5!=0&&a%7!=0)||(a%5==0&&a%3!=0&&a%7!=0)||(a%7==0&&a%5!=0&&a%3!=0))
{
printf("2222");

}
if(a%3!=0&&a%5!=0&&a%7!=0)
{
printf("3333");
// getch();
}
我用gcc编译没错了

Answer 2

很明显的错误，scanf("%d",&a); 一定要有双引号 ，不用getch（）；

Answer 3

拿去用！
void Test(int n)
{
if (n % 3 == 0 && n % 5 == 0 && n % 7 == 0)
{
cout << "Can be divided by 3, 5 and 7" << endl ;
}
else if (n % (3 * 5) == 0 && n % 7 != 0)
{
cout << "Can be divided by 3 and 5 but not 7!" << endl ;
}
else if (n % (3 * 7) == 0 && n % 5 != 0)
{
cout << "Can be divided by 3 and 7 but not 5!" << endl ;
}
else if (n % (5 * 7) == 0 && n % 3 != 0)
{
cout << "Can be divided by 5 and 7 but not 3!" << endl ;
}
else if (n % 3 == 0 && n % (5 * 7) != 0)
{
cout << "Can be divided by 3!" << endl ;
}
else if (n % 5 == 0 && n % (3 * 7) != 0)
{
cout << "Can be divided by 5!" << endl ;
}
else if (n % 7 == 0 && n % (3 * 5) != 0)
{
cout << "Can be divided by 7!" << endl ;
}
else
{
cout << "Can not be divided by any of 3, 5, 7!" << endl ;
}
}本回答被提问者采纳